How can I convert a String "Hello" to an Array ["H","e","l","l","o"] in Swift?
In Objective-C I have used this:
NSMutableArray *characters = [[NSMutableArray alloc] initWithCapacity:[myString length]];
for (int i=0; i < [myString length]; i++) {
NSString *ichar = [NSString stringWithFormat:@"%c", [myString characterAtIndex:i]];
[characters addObject:ichar];
}
It is even easier in Swift:
let string : String = "Hello 🐶🐮 🇩🇪"
let characters = Array(string)
println(characters)
// [H, e, l, l, o, , 🐶, 🐮, , 🇩🇪]
This uses the facts that
Array can be created from a SequenceType, andString conforms to the SequenceType protocol, and its sequence generator
enumerates the characters.And since Swift strings have full support for Unicode, this works even with characters outside of the "Basic Multilingual Plane" (such as 🐶) and with extended grapheme clusters (such as 🇩🇪, which is actually composed of two Unicode scalars).
Update: As of Swift 2, String does no longer conform to
SequenceType, but the characters property provides a sequence of the
Unicode characters:
let string = "Hello 🐶🐮 🇩🇪"
let characters = Array(string.characters)
print(characters)
This works in Swift 3 as well.
Update: As of Swift 4, String is (again) a collection of its
Characters:
let string = "Hello 🐶🐮 🇩🇪"
let characters = Array(string)
print(characters)
// ["H", "e", "l", "l", "o", " ", "🐶", "🐮", " ", "🇩🇪"]
let joined = ", ".join(characters);) the array results in an 'String' is not identical to 'Character' error?
", ".join(...) expects an array of strings. devxoul's answer below shows how you could split into an array of strings.characters is a collection and a view on the strings characters. Creating an array from the characters makes a copy. – Of course can access the characters directly: for i in string.characters.indices { print(string.characters[i]) } or just for c in string.characters { print(c) }In Swift 4, you don't have to use characters to use map(). Just do map() on String.
let letters = "ABC".map { String($0) }
print(letters) // ["A", "B", "C"]
print(type(of: letters)) // Array<String>
Or if you'd prefer shorter: "ABC".map(String.init) (2-bytes 😀)
In Swift 2 and Swift 3, You can use map() function to characters property.
let letters = "ABC".characters.map { String($0) }
print(letters) // ["A", "B", "C"]
Accepted answer doesn't seem to be the best, because sequence-converted String is not a String sequence, but Character:
$ swift
Welcome to Swift! Type :help for assistance.
1> Array("ABC")
$R0: [Character] = 3 values {
[0] = "A"
[1] = "B"
[2] = "C"
}
This below works for me:
let str = "ABC"
let arr = map(str) { s -> String in String(s) }
Reference for a global function map() is here: http://swifter.natecook.com/func/map/
arr = map(str) { String($0) }
Array initializer instead: Array("ABC".characters)
"ABC".characters and Array("ABC".characters)."ABC".characters returns a type String.CharacterView. To convert it to an array, specifically of type, [Character], you'd have to use the Array initializer on it.[String] instead of [Character] because op wanted a string array :)There is also this useful function on String: components(separatedBy: String)
let string = "1;2;3"
let array = string.components(separatedBy: ";")
print(array) // returns ["1", "2", "3"]
Works well to deal with strings separated by a character like ";" or even "\n"
For Swift version 5.3 its easy as:
let string = "Hello world"
let characters = Array(string)
print(characters)
// ["H", "e", "l", "l", "o", " ", "w", "o", "r", "l", "d"]
Updated for Swift 4
Here are 3 ways.
//array of Characters
let charArr1 = [Character](myString)
//array of String.element
let charArr2 = Array(myString)
for char in myString {
//char is of type Character
}
In some cases, what people really want is a way to convert a string into an array of little strings with 1 character length each. Here is a super efficient way to do that:
//array of String
var strArr = myString.map { String($0)}
Swift 3
Here are 3 ways.
let charArr1 = [Character](myString.characters)
let charArr2 = Array(myString.characters)
for char in myString.characters {
//char is of type Character
}
In some cases, what people really want is a way to convert a string into an array of little strings with 1 character length each. Here is a super efficient way to do that:
var strArr = myString.characters.map { String($0)}
Or you can add an extension to String.
extension String {
func letterize() -> [Character] {
return Array(self.characters)
}
}
Then you can call it like this:
let charArr = "Cat".letterize()
An easy way to do this is to map the variable and return each Character as a String:
let someText = "hello"
let array = someText.map({ String($0) }) // [String]
The output should be ["h", "e", "l", "l", "o"].
for the function on String: components(separatedBy: String)
in Swift 5.1
have change to:
string.split(separator: "/")
Martin R answer is the best approach, and as he said, because String conforms the SquenceType protocol, you can also enumerate a string, getting each character on each iteration.
let characters = "Hello"
var charactersArray: [Character] = []
for (index, character) in enumerate(characters) {
//do something with the character at index
charactersArray.append(character)
}
println(charactersArray)
for character in characters { charactersArray.append(character)} without the enumeration() let string = "hell0"
let ar = Array(string.characters)
print(ar)
In Swift 4, as String is a collection of Character, you need to use map
let array1 = Array("hello") // Array<Character>
let array2 = Array("hello").map({ "\($0)" }) // Array<String>
let array3 = "hello".map(String.init) // Array<String>
Suppose you have four text fields otpOneTxt, otpTwoTxt, otpThreeTxt, otpFourTxt and a string getOtp.
let getup = "5642"
let array = self.getOtp.map({ String($0) })
otpOneTxt.text = array[0] //5
otpTwoTxt.text = array[1] //6
otpThreeTxt.text = array[2] //4
otpFourTxt.text = array[3] //2
You can also create an extension:
var strArray = "Hello, playground".Letterize()
extension String {
func Letterize() -> [String] {
return map(self) { String($0) }
}
}
func letterize() -> [Character] {
return Array(self.characters)
}
let str = "cdcd"
let characterArr = str.reduce(into: [Character]()) { result, letter in
result.append(letter)
}
print(characterArr)
//["c", "d", "c", "d"]
